tag:blogger.com,1999:blog-5303307482158922565.post9009244238957163987..comments2024-03-22T13:39:55.941-07:00Comments on Math Mama Writes...: Pythagorean TriplesSue VanHattumhttp://www.blogger.com/profile/10237941346154683902noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-5303307482158922565.post-39773168995145228992012-12-02T07:57:14.475-08:002012-12-02T07:57:14.475-08:00I never did write out my way of doing it. Partly b...I never did write out my way of doing it. Partly because I wanted people to think it through. But the conventional procedure is everywhere. Hmm... New post?Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-30165064321318809542012-08-13T05:59:33.097-07:002012-08-13T05:59:33.097-07:00If I ever get back to playing with Pythagorean Tri...If I ever get back to playing with Pythagorean Triples more, this looks like a cool way to play with it: http://girlsangle.wordpress.com/2012/08/13/generating-pythagorean-triples-using-matrices/<br /><br />But this method produces the answers first, before we know why.Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-32753311512993854312012-08-06T09:35:43.882-07:002012-08-06T09:35:43.882-07:00I refer to Sue Van Hattum's comment of 19 July...I refer to Sue Van Hattum's comment of 19 July 2012. My variables of p and q are not really extra they are a different way of expressing the usual variables of a and b, and can be applied in the following proof of Fermat's Last Theorem. With p, q, x and n all assumed to be positive integers, and x less than or equal to n, each term of the binomial expansion series of (p+q)^n-(p-q)^n contains p^x.q^(n-x), so that if p equals q, then p^n equalling q^n becomes the common factor of all the terms in the binomial expansion series. The consequence is that the other factors in the terms all add up to 2^n, for example 6p^2.q^1 +2q^3 equals 2^3.p^3 equalling 2^3.q^3. This cannot happen without p equalling q except when n equals 2, and there is consequently only one term 2^2.pq, with both p and q unequal but having integer square roots, this is the Pythagorean Triple. However if unequal p and q do not have integer square roots then the power n equalling 2 is treated the same way as n being greater than 2, that is (p+q)^n -(p-q)^n cannot have an integer nth root unless p equals q, that is Fermat's Last Theorem. You have to know the binomial formula to follow this.Peter L. Griffithsnoreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-91313606915302313022012-07-19T08:39:56.920-07:002012-07-19T08:39:56.920-07:00Why 'best', Peter? (I see how it works, bu...Why 'best', Peter? (I see how it works, but the way I analyzed it, I didn't use extra variables. I like my way, and so I'm curious about the benefits of other methods.)Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-12358676732187565572012-07-19T07:52:56.235-07:002012-07-19T07:52:56.235-07:00The best way to identify Pythagorean Triples is to...The best way to identify Pythagorean Triples is to express the equation as (p+q)^2-(p-q)^2, so that the Pythagorean Triple arises when pq has an integer square root.Peter L. Griffithsnoreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-25602623509326463112010-09-01T20:02:31.634-07:002010-09-01T20:02:31.634-07:00Hi Sue,
I have developed a theorem that is based ...Hi Sue,<br /><br />I have developed a theorem that is based on a matrix of indeces (i,j) where i and j are natural numbers and i is odd and j is even. For every pair of indeces (i,j) where i and j are relatively prime, a unique pythagorean triangle is described. If you are interested further, you can e-mail me at martinb@apiu.edu<br /><br />Regards<br /><br />Martin BredenkampAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-13495685351605188152010-01-08T18:34:01.941-08:002010-01-08T18:34:01.941-08:00I think I remember the excitement part, but the re...I think I remember the excitement part, but the rest is definitely gone. ;^) I'll have to print out what you wrote so I can think about it properly. And I better print out the rest while I'm at it...<br /><br />(My son has a friend over and they're playing chess, so maybe it's the perfect time to do a math problem. But the way they play chess is not quite civilized - I might have to intervene at any moment. I'm not sure how many rules they know yet...)Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-77004837015990783642010-01-08T15:25:47.825-08:002010-01-08T15:25:47.825-08:00Hey Sue! Happy new year!
I loved that problem too...Hey Sue! Happy new year!<br /><br />I loved that problem too this summer. <a href="http://mathbebrave.blogspot.com" rel="nofollow">Jesse</a> and I have actually led this problem as a math circle 3 times this fall. (All three times for groups of teachers.)<br /><br />The train of thought Bob and Ellen led us on actually proved that all ppt's have the form Vlorbik and Curmudgeon described. I remember you contributing a key insight this summer when we did the problem with the Kaplans.<br /><br />SPOILER ALERT<br /><br />We had shown early on that in a ppt, not only do the three a,b,c lack a common factor but each pair (for example a,c) also has no common factor. We had also already shown as a group that c has to be odd and that one of a,b was even. We agreed to call the even one b (some of us were a little sad to give up our attachment to the convention that a was the smallest, but bigger things were afoot...) - then we wrote:<br /><br />b^2 = c^2 - a^2<br />b^2 = (c+a)(c-a)<br /><br />Bob asked us if c+a and c-a had any common factors. We all agreed 2 was a common factor because a and c are both odd so c+a and c-a are both even. Then Bob said, "the next insight - which I have never figured out how to motivate - is that this is their <i>only</i> common factor." (Jesse and I have experimented with trying to get people to come to this on their own by asking questions like "how far apart are c-a and c+a? does this tell you anything about their common factors?" This has been the hardest step for every single group we've done it with.)<br /><br />With Bob & Ellen, the proof that a-c and a+c only have 2 as a common factor was done by first dividing out the 2 from both factors of each side:<br /><br />(b/2)^2 = [(c+a)/2][(c-a)/2]<br /><br />(b is even, and a-c and a+c are both even, so everything is still an integer). Then we noted that the sum of the two factors on the right is c while the difference is a. Thus if (c+a)/2 and (c-a)/2 had a common factor, it would also be a common factor of c and a; but we had decided early on that they don't have any. So neither do (c+a)/2 and (c-a)/2.<br /><br />At this point I distinctly remember you, Sue Vanhattum, getting very excited and saying something to the mathematical effect of, "But wait... the left side (b/2)^2 is a square! If you look at its prime factorization, all the factors will come in pairs. But the two factors on the right side, (c+a)/2 and (c-a)/2, don't have any common factors! This means that they can't each contribute one half of a pair of identical factors, because then they would have a common factor. But this means that each of their factorizations is also composed of complete pairs, which means they are both square!"<br /><br />Bob put your insight into algebraic language: (c+a)/2 = u^2; (c-a)/2 = v^2.<br /><br />The rest was just playing out the consequences:<br /><br />a = (c+a)/2 - (c-a)/2 = u^2 - v^2<br />c = (c+a)/2 + (c-a)/2 = u^2 + v^2<br /><br />and<br /><br />(b/2)^2=[(c+a)/2][(c-a)/2] = u^2*v^2<br /><br />so<br /><br />b/2 = uv, i.e.<br /><br />b = 2uv<br /><br />and the ppt turned out to be exactly what vlorbik and curmudgeon said.<br /><br />(At this point this summer we had shown that all ppts have this form; we didn't show that anything with this form is a ppt. Vlorbik's calculation shows it is a pt for sure. Some slight additional conditions, which Vlorbik mentions, need to be put on u and v to guarantee that u^2-v^2, 2uv, u^2+v^2 is primitive. I had fun convincing myself that these conditions are necessary and sufficient.)<br /><br />Btw, if you're interested in a hint in showing that one of the three members is divisible by 5, then here is one:<br /><br />Consider u's and v's remainders when divided by 5; then, what are the possible remainders of u^2 and v^2 when divided by 5?Ben Blum-Smithhttp://researchinpractice.wordpress.comnoreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-1832964248976274472009-12-30T07:33:27.370-08:002009-12-30T07:33:27.370-08:00pick integer values of u and v, where u > v.
f...pick integer values of u and v, where u > v.<br /><br />find the three values<br />u^2 + v^2<br />u^2 - v^2<br />2uv<br /><br />These will be pythagorean triples in some order. Which one is the hypotenuse is trivial.Curmudgeonhttps://www.blogger.com/profile/04323026187622872114noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-88899441881207622052009-12-30T05:36:46.659-08:002009-12-30T05:36:46.659-08:00I love that all 3 of these comments describe diffe...I love that all 3 of these comments describe different ways of finding PPT's.<br /><br />@ Calvin: I think I already gave the infinite number question away in the post. It's cool.<br /><br />@ Glenn and vlorbik: I love that fibonacci numbers and complex numbers come into this - everything is connected.<br /><br />What you're all making clearer to me is that I described a procedure different from the one I saw at the Institute. So I'm coming up with something new (to me). I like that.<br /><br />After I finished writing this, I went to Wikipedia. The article on Pythagorean triples made dozens of claims that weren't proven. If you like playing with this and want more to prove than what you can conjecture on your own, it's a treasure.<br /><br />Part of what I like about this problem is that one needs very little background material for it. <br /><br />I am so grateful to the Kaplans and to the internet for getting me doing math again. It's a joy for me. :^)Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-71918622015525987772009-12-30T04:44:08.414-08:002009-12-30T04:44:08.414-08:00pick a > b coprime, with one even.
x= a^2 + b^...pick a > b coprime, with one even.<br /><br />x= a^2 + b^2<br /><br />y = 2ab<br /><br />then x^2 + y^2 =<br />a^4 - 2a^2b^2 + b^4 + 4a^2b^2 =<br />(a^2+b^2)^2<br />so (x, y, a^2+b^2)<br />is a ppt.<br /><br />i saw a student present this trick<br />at capital u when i'd forgotten it.<br />never forgot it since.<br /><br />it's worth remarking in this context<br />(repeatedly) that since the complex<br />norm... magnitude, length, abs...<br />is the sum of the squares of the<br />real & imaginary parts<br />abs(x + i y) = x^2 + y^2<br />and the norm is "multiplicative"<br />( abs(zw) = abs(z) abs(w) )<br />one sees by brute calculation<br />that the *product* of <br />sums-of-two-squares<br />is itself a sum-of-two-squares.<br /><br />i studied those in another life.r. r. vlorbikhttps://www.blogger.com/profile/02746118913980983815noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-5163091344283910832009-12-30T01:54:32.454-08:002009-12-30T01:54:32.454-08:00Hi Sue,
might be a spoiler....
did you know this ...Hi Sue,<br /><br />might be a spoiler....<br />did you know this one (picked up from the book 'Number' by Tobias Dantzig):<br /><br />For any value of u and v such that 2uv is a perfect square we obtain a Pythagorean triangle.<br />x = u + sqrt(2uv)<br />y = v + sqrt(2uv)<br />z = u + v + sqrt(2uv)<br /><br />that answers your question: there are infinitely many!Calvinhttps://www.blogger.com/profile/00330395410757657740noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-65566806428839599522009-12-29T22:09:33.328-08:002009-12-29T22:09:33.328-08:00I know that you can generate an infinite number of...I know that you can generate an infinite number of pythagorean triples using fibonacci numbers. I know of two different ways to do this, but here is a link that shows one method. http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#pythfib <br /><br />The second method is hinted at at the end of the paragraph. In my lesson on fibonacci numbers, I have my learners use one method to calculate a triple, and then find the two fib. numbers using a second method that will create the same triple.<br /><br />It is fairly cool to do, actually.Glennhttps://www.blogger.com/profile/15228515637649790365noreply@blogger.com